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How can we prove that the space of continuous functions with compact support is dense in the space of continuous functions that vanish at infinity?

akansha
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    Please show some effort in your questions. You have asked 3 questions in the space of 20 minutes all without any effort or courtesy. – Vishal Gupta Aug 11 '13 at 18:56
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    This question is by no means off-topic (though I think it has nothing to do with functional analysis. Neither does it have to hold in metric space (LCH is ok)). I also think the question was very clearly presented (one sentence is enough). Maybe there is duplication because a user mentioned that above. I also agree that it is not good if the asker showed no effort. But it is unreasonable to close this question under an incorrect reason. As for the answer to this question, it is Proposition 4.35 in P132 of text "Real Analysis: Modern Techniques and Their Applications" by Gerald B. Folland, ... – user5280911 Aug 30 '20 at 14:53
  • ... proved in LCH. – user5280911 Aug 30 '20 at 14:53

1 Answers1

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Presumably your norm is the maximum absolute distance at any point. If I give you a continuous $f$ that vanishes at $\infty$ and an $\epsilon \gt 0$, you need to find a continuous $g$ with compact support that is everywhere within $\epsilon$ of $f$. Since $f$ vanishes at $\infty$, there is some $N$ such that $f(x) \lt \epsilon/2$ whenever $ x \gt N$. So if $g=f$ out to $N$, drops quickly to $0$ and stays there...

Added: We are trying to show that any continuous function that vanishes at infinity can be approximated closely (within $\epsilon$) by a function with compact support. The range where $f$ is large (greater that $\epsilon/2$) is some interval $(M,N)$ Our approximating function will agree with $f$ over that range. We then know that $|f(x)| \lt \epsilon/2$ on $(-\infty,M)$ and $(N,\infty)$. If we let the approximating function go to zero on $(M',M)$ and $(N,N')$ and be zero on $(-\infty,M')$ and $(N',\infty)$ it will have compact support and be within $\epsilon$ of $f$ over the whole line.

Ross Millikan
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  • If the domain is not the space of real numbers then how to maintain continuity of the function. – akansha Aug 12 '13 at 12:02
  • What is your domain and range? When I saw compact support and vanish at infinity I thought $\Bbb R \to \Bbb R$, but it works in $\Bbb C$ as well. – Ross Millikan Aug 12 '13 at 13:48
  • If I take domain X to be any topological space and the space of all continuous real valued functions on X with compact support with sup norm. Then, what will be its completion and how ? – akansha Aug 12 '13 at 16:58
  • The same approach works as long as you have a similar concept of "go to zero at infinity". You need to have the function value to be forced to be small as long as it is "big enough". Of course, if $X$ is compact, you don't have this concept, but functions with compact support are not so special then. – Ross Millikan Aug 12 '13 at 19:55
  • Could you explain this again? I am confused about the N that you need. – Tyler Hilton Sep 29 '13 at 21:48
  • @TylerHilton: see if this helps – Ross Millikan Sep 29 '13 at 22:35
  • Will $g$ still be in the set of functions with compact support in $\mathbb{R}$ when $f(x)=0$ for some $x\in [-N,N]$? – Kurome Mar 06 '17 at 12:54
  • @Kurome: you are right, if $f=0$ at a point then the $g$ I defined would not have compact support. It can be fixed by defining $g$ to be slightly nonzero. – Ross Millikan Mar 06 '17 at 16:28
  • Could you confirm that result does not hold in general, but only on LCH spaces? – Akira May 11 '22 at 21:21